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Unformatted text preview: CHEE 3363Fluid Mechanics HW 4 Solutions Pro 4.12 Given: Control volume with linear velocity distribution across surface ① as shown; width = ? . Find: (a) Volume flow rate (b) Momentum flux, through surface ① Solutions: The volume flow rate is ¡ = ? ¢£ ∙ ¤ £ At surface ① ? ¢£ = ? ℎ ?¥¦ ¤ = −§¤?¥¦ Thus ¡ = ? ℎ ?¥¦ ∙ ( −§¤?¥¦ ) ℎ ? =0 = − ?§ ℎ ?¤? ℎ = − ?§ ℎ ¨ ? 2 2 © ℎ ¡ = − 1 2 ?ℎ§ The momentum flux is given by ª . « . = ? ¢£ ( ¬? ¢£ ∙ ¤ £ ) Thus ª . « . = ? ℎ ?¥¦ −¬ ?§ ℎ ?¤?® = −¬ ? 2 § ℎ 2 ¥¦ ? 2 ¤? ℎ = −¬ ? 2 § ℎ 2 ¥¦ ¨ ? 3 3 © ℎ ℎ ª . « . = − 1 3 ¬? 2 §ℎ¥¦ Pro 4.15 Given: Velocity distribution for laminar flow in a long circular tube ? ¡ = ??¢ = ? £¤¥ ¦ 1 − § ¨ © ª 2 « ?¢ where © is the tube radius Evaluate: (a) The volume flow rate (b) The momentum flux through a section normal to the pipe axis Solution: The volume flow rate is given by...
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 Spring '11
 Sarkozi
 Fluid Dynamics, Fluid Mechanics, Flux, Incompressible Flow, volume flow rate

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